Denote the area of any triangle $\triangle \Delta$ by $S_{\triangle \Delta}$. Let $D'$ be the reflection of $D$ from $AC$ and $\alpha = \angle APB$ (Fig. 24). We have

$$S_{PAB} = \frac{1}{2} \cdot PA \cdot PB \cdot \sin \alpha, \qquad (4)$$
$$S_{PBC} = \frac{1}{2} \cdot PB \cdot PC \cdot \sin(180^\circ - \alpha) = \frac{1}{2} \cdot PB \cdot PC \cdot \sin \alpha, \quad (5)$$
$$S_{PCD} = \frac{1}{2} \cdot PC \cdot PD \cdot \sin \alpha, \qquad (6)$$
$$S_{PDA} = \frac{1}{2} \cdot PD \cdot PA \cdot \sin(180^\circ - \alpha) = \frac{1}{2} \cdot PD \cdot PA \cdot \sin \alpha. \quad (7)$$

Fig. 24

$\angle D'AC < \angle BAC$. By interchanging the roles of $A$ and $C$, we similarly obtain $\angle D'CA < \angle BCA$. Hence $D'$ lies inside the triangle $ABC$, implying that $S_{ADC} = S_{AD'C} < S_{ABC}$. Adding (6) and (7) gives $S_{PCD} + S_{PDA} = \frac{1}{2} \cdot (PC + PA) \cdot PD \cdot \sin \alpha$, i.e., $S_{ACD} = \frac{1}{2} \cdot AC \cdot PD \cdot \sin \alpha$. Analogously from (4) and (5) we obtain $S_{ABC} = \frac{1}{2} \cdot AC \cdot PB \cdot \sin \alpha$. Hence the inequality $S_{ADC} < S_{ABC}$ implies $PD < PB$.

Interchanging the roles of $A$ and $B$ and the roles of $C$ and $D$ similarly gives $PC < PA$. Now (4) and (5) together imply $S_{PAB} > S_{PBC}$, (4) and (7) together imply $S_{PAB} > S_{PDA}$, (5) and (6) together imply $S_{PBC} > S_{PCD}$, and (6) and (7) together imply $S_{PDA} > S_{PCD}$. Hence the desired claim follows.